I am studying undergraduate complex analysis, and in my Textbook the author claimed that
$$\arctan(2)=\frac{\pi}{2}-\arctan\left(\frac{1}{2}\right)$$
when he was doing an example regarding to principle argument. The origin of his claim is definitely not layed in his book so I was wondering if there is a rigorous proof that can explain this equality.
I have tried to prove this statement using Taylor series but I just can't reduce the term to the equation above.
Any hint would be much appreciated
$\endgroup$ 24 Answers
$\begingroup$Let $f:(0,\infty)\to\mathbb R$ be defined by $f(x)=\arctan x+\arctan \frac1x$. Then $$f'(x)=\dfrac{1}{1+x^2}+\dfrac{1}{1+\left(\frac{1}{x}\right)^2}\cdot\dfrac{-1}{x^2}=0$$ for all $x>0$. Hence $f$ is constant on $(0,\infty)$ and $f(x)=f(1)=\frac\pi2$ for all $x>0$, including $x=2$.
$\endgroup$ $\begingroup$Note that $$ \tan(\pi/2-x)=\frac{\sin(\pi/2-x)}{\cos(\pi/2-x)}. $$ But $\sin(\pi/2-x)=\cos(x)$ and $\cos(\pi/2-x)=\sin(x)$. So, we have $$ \tan(\pi/2-x)=\frac{\cos(x)}{\sin(x)}=\frac{1}{\tan(x)}. $$ So, if you take $$ \tan\left(\frac{\pi}{2}-\arctan(2)\right)=\frac{1}{\tan(\arctan(2))}=\frac{1}{2}. $$ Since $-\frac{\pi}{2}<\frac{\pi}{2}-\arctan(2)<\frac{\pi}{2}$, this implies that $$ \frac{\pi}{2}-\arctan(2)=\arctan\left(\frac{1}{2}\right), $$ as claimed.
$\endgroup$ $\begingroup$If you want a proof without (overt) appeal to geometry, you can observe that $\arctan 2$ is the principal argument of $1+2i$, and $\arctan \frac 12$ is the principal argument of $1+\frac12 i$. Therefore their sum is an argument of $(1+2i)(1+\frac12 i) = \frac52 i$, so it must be $\pi/2+2\pi n$ for some $n$. But by considering the range of $\arctan$ we can see that $n=0$ is the only possibility, so $$ \arctan 2 + \arctan\frac12 = \frac\pi2 $$
$\endgroup$ $\begingroup$In fact arctanx=pi/2-arctan1/x, there are few ways of proving this, one way is the right-angled triangle way suggested by user84413 who just commented on your post, or you could tan both sides and carefully show they are both equivalent.
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